We are given a random collection of 3 cards. (a) What is the conditional probability that all 3 cards are spades, given that the collection contains the ace of spades? Find an approximate decimal. (b) What is the conditional probability that all 3 cards are spades, given that at least one card is a spade? Find an approximate decimal.

a) Assuming a standard deck of 52 cards. By definition of conditional probability:

P(3 spades | ace of spades) = P(3 spades AND ace of spades) / P(ace of spades)

We draw 3 cards, so there are (52 choose 3) = 22,100 total possible hands.

There are (1 choose 1)*(51 choose 2) = 1275 possible hands with the ace of spades.

There are (13 choose 3) = 286 possible hands containing 3 spades, and (1 choose 1)*(12 choose 2) = 66 of these contain the ace of spades.

So we have

P(3 spades AND ace of spades) = 66/22,100

P(ace of spades = 1275/22,100

and so

P(3 spades | ace of spades) = 66/1275 = 0.0518

b) By definition:

P(3 spades | at least 1 spade) = P(3 spades AND at least 1 spade) / P(at least 1 spade)

If you draw 3 spades, you already meet the requirement of having at least 1 spade, so

P(3 spades AND at least 1 spade) = P(3 spades)

and we already know there are 286 possible such hands.

The event of getting at least 1 spades is complementary to the event of not getting any spades, meaning

P(at least 1 spade) = 1 - P(no spades)

There are 39 non-spade cards, and so (39 choose 3) = 9139 possible hands not containing spades.

So we find

P(3 spades) = 286/22,100

P(at least 1 spade) = 1 - 9139/22,100 = 12,961/22,100

which means

P(3 spades | at least 1 spade) = 286/12,961 = 0.0221

no

step-by-step explanation:

2-7=-5

7-2=5

when you combing like terms for ech expression, 5 and -5. these numbers are not equal to each other.

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09